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5 Vector Space \mathbb{R}^n

5.1 Subspaces and Spanning

In Section 2.2 we introduced the set \mathbb{R}^n of all n-tuples (called \textit{vectors}), and began our investigation of the matrix transformations \mathbb{R}^n \to \mathbb{R}^m given by matrix multiplication by an m \times n matrix. Particular attention was paid to the euclidean plane \mathbb{R}^2 where certain simple geometric transformations were seen to be matrix transformations.

In this chapter we investigate \mathbb{R}^n in full generality, and introduce some of the most important concepts and methods in linear algebra. The n-tuples in \mathbb{R}^n  will continue to be denoted \vec{x}, \vec{y}, and so on, and will be written as rows or columns depending on the context.

Subspaces of \mathbb{R}^n

Definition 5.1 Subspaces of \mathbb{R}^n

A set U of vectors in \mathbb{R}^n is called a subspace of \mathbb{R}^n if it satisfies the following properties:

S1. The zero vector \vec{0} \in U.

S2. If \vec{x} \in U and \vec{y} \in U, then \vec{x} + \vec{y} \in U.

S3. If \vec{x} \in U, then a\vec{x} \in U for every real number a.

We say that the subset U is closed under addition if S2 holds, and that U is closed under scalar multiplication if S3 holds.

Clearly \mathbb{R}^n is a subspace of itself, and this chapter is about these subspaces and their properties. The set U = \{\vec{0}\}, consisting of only the zero vector, is also a subspace because \vec{0} + \vec{0} = \vec{0} and a\vec{0} = \vec{0} for each a in \mathbb{R} ; it is called the zero subspace. Any subspace of \mathbb{R}^n other than \{\vec{0}\} or \mathbb{R}^n is called a proper subspace.

 

 

A planeWe saw in Section 4.2 that every plane M through the origin in \mathbb{R}^3 has equation ax + by + cz = 0 where a, b, and c are not all zero. Here \vec{n} = \left[ \begin{array}{r} a \\ b \\ c \end{array} \right]  is a normal for the plane and

    \begin{equation*} M = \{ \vec{v} \mbox{ in }\mathbb{R}^3 \mid \vec{n} \cdot \vec{v} = 0 \} \end{equation*}

where
\vec{v} = \left[ \begin{array}{r} x \\ y \\ z \end{array} \right] and \vec{n} \cdot \vec{v} denotes the dot product introduced in Section 2.2 (see the diagram). Then M is a subspace of \RR^3. Indeed we show that M satisfies S1, S2, and S3 as follows:

S1. \vec{0} \in M because \vec{n} \cdot \vec{0} = 0;

S2. If \vec{v} \in M and \vec{v}_{1} \in M, then \vec{n} \cdot (\vec{v} + \vec{v}_{1}) = \vec{n} \cdotprod \vec{v} + \vec{n} \cdot \vec{v}_{1} = 0 + 0 = 0, so \vec{v} + \vec{v}_{1} \in M;

S3. If \vec{v} \in M, then \vec{n} \cdot (a\vec{v}) = a(\vec{n} \cdot \vec{v}) = a(0) = 0, so a\vec{v} \in M.

Example 5.1.1

Planes and lines through the origin in \mathbb{R}^3 are all subspaces of \mathbb{R}^3.

Solution:

We proved the statement for planes above. If L is a line through the origin with direction vector \vec{d}, then L = \{t\vec{d} \mid t \in \mathbb{R}\}. Can you verify that L satisfies S1, S2, and S3?

Example 5.1.1 shows that lines through the origin in \mathbb{R}^2 are subspaces; in fact, they are the only proper subspaces of \mathbb{R}^2. Indeed, we will prove that lines and planes through the origin in \mathbb{R}^3 are the only proper subspaces of \mathbb{R}^3. Thus the geometry of lines and planes through the origin is captured by the subspace concept. (Note that every line or plane is just a translation of one of these.)

 

 

Subspaces can also be used to describe important features of an m \times n matrix A. The null space of A, denoted \func{null} A, and the image space of A, denoted \func{im} A, are defined by

    \begin{equation*} \func{null } A = \{ \vec{x} \in \mathbb{R}^n \mid A\vec{x} = \vec{0}\} \quad \mbox{ and } \quad \func{im } A = \{A\vec{x} \mid \vec{x} \in \mathbb{R}^n\} \end{equation*}

In the language of Chapter 2, \func{null} A consists of all solutions \vec{x} in \mathbb{R}^n of the homogeneous system A\vec{x} = \vec{0}, and \func{im} A is the set of all vectors \vec{y} in \mathbb{R}^m such that A\vec{x} = \vec{y}  has a solution \vec{x}. Note that \vec{x} is in \func{null} A if it satisfies the condition A\vec{x} = \vec{0}, while \func{im} A consists of vectors of the form A\vec{x} for some \vec{x} in \mathbb{R}^n. These two ways to describe subsets occur frequently.

Example 5.1.2

If A is an m \times n matrix, then:

  1. \func{null} A is a subspace of \mathbb{R}^n.
  2. \func{im} A is a subspace of \mathbb{R}^m.

Solution:

  1. The zero vector \vec{0} \in \mathbb{R}^n lies in \func{null} A because A\vec{0} = \vec{0}. If \vec{x} and \vec{x}_{1} are in \func{null} A, then \vec{x} + \vec{x}_{1} and a\vec{x} are in \func{null} A because they satisfy the required condition:

        \begin{equation*} A(\vec{x} + \vec{x}_1) = A\vec{x} + A\vec{x}_1 = \vec{0} + \vec{0} = \vec{0} \quad \mbox{and} \quad A(a\vec{x}) = a(A\vec{x}) = a\vec{0} = \vec{0} \end{equation*}

    Hence \func{null} A satisfies S1, S2, and S3, and so is a subspace of \mathbb{R}^n.

  2. The zero vector \vec{0} \in \mathbb{R}^m lies in \func{im} A because \vec{0} = A\vec{0}. Suppose that \vec{y} and \vec{y}_{1} are in \func{im} A, say \vec{y} = A\vec{x} and \vec{y}_{1} = A\vec{x}_{1} where \vec{x} and \vec{x}_{1} are in \mathbb{R}^n. Then

        \begin{equation*} \vec{y} + \vec{y}_1 = A\vec{x} + A\vec{x}_1 = A(\vec{x} + \vec{x}_1) \quad \mbox{and} \quad a\vec{y} = a(A\vec{x}) = A(a\vec{x}) \end{equation*}

    show that \vec{y} + \vec{y}_{1} and a\vec{y} are both in \func{im} A (they have the required form). Hence \func{im} A is a subspace of \mathbb{R}^m.

There are other important subspaces associated with a matrix A that clarify basic properties of A. If A is an n \times n matrix and \lambda is any number, let

    \begin{equation*} E_\lambda(A) = \{\vec{x} \in \mathbb{R}^n \mid A\vec{x} = \lambda\vec{x} \} \end{equation*}

A vector \vec{x} is in E_\lambda(A) if and only if (\lambda I - A)\vec{x} = \vec{0}, so Example 5.1.2 gives:

Example 5.1.3

E_\lambda(A) = \func{null}(\lambda I- A) is a subspace of \mathbb{R}^n for each n \times n matrix A and number \lambda.

E_\lambda(A) is called the eigenspace of A corresponding to \lambda. The reason for the name is that, in the terminology of Section 3.3, \lambda is an eigenvalue of A if E_\lambda(A) \neq \{\vec{0}\}. In this case the nonzero vectors in E_\lambda(A) are called the eigenvectors of A corresponding to \lambda.

The reader should not get the impression that every subset of \mathbb{R}^n is a subspace. For example:

    \begin{align*} U_1 &= \left \{ \left[ \begin{array}{l} x \\ y \end{array} \right] \, \middle| \, x \geq 0 \right\} \mbox{ satisfies S1 and S2, but not S3;} \\ U_2 &= \left \{ \left[ \begin{array}{l} x \\ y \end{array} \right] \, \middle| \, x^2 = y^2 \right\} \mbox{ satisfies S1 and S3, but not S2;} \end{align*}

Hence neither U_{1} nor U_{2} is a subspace of \RR^2.

 

 

Spanning sets

Let \vec{v} and \vec{w} be two nonzero, nonparallel vectors in \mathbb{R}^3 with their tails at the origin. The plane M through the origin containing these vectors is described in Section 4.2 by saying that \vec{n} = \vec{v} \times \vec{w} is a normal for M, and that M consists of all vectors \vec{p} such that \vec{n} \cdot\vec{p} = 0.
While this is a very useful way to look at planes, there is another approach that is at least as useful in \mathbb{R}^3 and, more importantly, works for all subspaces of \mathbb{R}^n for any n \geq 1.

Spanning setThe idea is as follows: Observe that, by the diagram, a vector \vec{p} is in M if and only if it has the form

    \begin{equation*} \vec{p} = a\vec{v} + b\vec{w} \end{equation*}

for certain real numbers a and b (we say that \vec{p} is a linear combination of \vec{v} and \vec{w}).
Hence we can describe M as

    \begin{equation*} M = \{a\vec{x} + b\vec{w} \mid a, b \in \mathbb{R} \} \end{equation*}

and we say that \{\vec{v}, \vec{w}\} is a spanning set for M. It is this notion of a spanning set that provides a way to describe all subspaces of \mathbb{R}^n.

As in Section 1.3, given vectors \vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k} in \mathbb{R}^n, a vector of the form

    \begin{equation*} t_1\vec{x}_1 + t_2\vec{x}_2 + \dots + t_k\vec{x}_k \quad \mbox{where the } t_i \mbox{ are scalars} \end{equation*}

is called a linear combination of the \vec{x}_{i}, and t_{i} is called the coefficient of \vec{x}_{i} in the linear combination.

Definition 5.2 Linear Combinations and Span in \mathbb{R}^n

The set of all such linear combinations is called the span of the \vec{x}_{i} and is denoted

    \begin{equation*} \func{span}\{\vec{x}_1, \vec{x}_2, \dots, \vec{x}_k\} = \{t_1\vec{x}_1 + t_2\vec{x}_2 + \dots + t_k\vec{x}_k \mid t_i \mbox{ in } \mathbb{R}\} \end{equation*}

If V = \func{span}\{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\}, we say that V is spanned by the vectors \vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}, and that the vectors \vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k} span the space V.

Here are two examples:

    \begin{equation*} \func{span}\{ \vec{x} \} = \{t\vec{x} \mid t \in \mathbb{R}\} \end{equation*}

which we write as \func{span}\{\vec{x}\} = \mathbb{R} \vec{x} for simplicity.

    \begin{equation*} \func{span}\{\vec{x}, \vec{y}\} = \{r\vec{x} + s\vec{y} \mid r, s \in \mathbb{R} \} \end{equation*}

In particular, the above discussion shows that, if \vec{v} and \vec{w} are two nonzero, nonparallel vectors in \mathbb{R}^3, then

    \begin{equation*} M = \func{span}\{\vec{v}, \vec{w} \} \end{equation*}

is the plane in \mathbb{R}^3 containing \vec{v} and \vec{w}. Moreover, if \vec{d} is any nonzero vector in \mathbb{R}^3 (or \mathbb{R}^2), then

    \begin{equation*} L = \func{span} \{\vec{v}\} = \{t\vec{d} \mid t \in \mathbb{R}\} = \mathbb{R}\vec{d} \end{equation*}

is the line with direction vector \vec{d}. Hence lines and planes can both be described in terms of spanning sets.

Example 5.1.4

Let \vec{x} = (2, -1, 2, 1) and \vec{y} = (3, 4, -1, 1) in \mathbb{R}^4. Determine whether \vec{p} = (0, -11, 8, 1) or \vec{q} = (2, 3, 1, 2) are in U = \func{span}\{\vec{x}, \vec{y}\}.

Solution:

The vector \vec{p} is in U if and only if \vec{p} = s\vec{x} + t\vec{y} for scalars s and t. Equating components gives equations

    \begin{equation*} 2s + 3t = 0, \quad -s + 4t = -11, \quad 2s - t = 8, \quad \mbox{and} \quad s + t = 1 \end{equation*}

This linear system has solution s = 3 and t = -2, so \vec{p} is in U. On the other hand, asking that \vec{q} = s\vec{x} + t\vec{y} leads to equations

    \begin{equation*} 2s + 3t = 2, \quad -s +4t = 3, \quad 2s -t = 1, \quad \mbox{and} \quad s + t = 2 \end{equation*}

and this system has no solution. So \vec{q} does not lie in U.

 

 

Theorem 5.1.1

Let U = \func{span}\{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} in \mathbb{R}^n. Then:

  1. U is a subspace of \mathbb{R}^n containing each \vec{x}_{i}.
  2. If W is a subspace of \mathbb{R}^n and each \vec{x}_{i} \in W, then U \subseteq W.

Proof:

  1. The zero vector \vec{0} is in U because \vec{0} = 0\vec{x}_{1} + 0\vec{x}_{2} + \dots + 0\vec{x}_{k} is a linear combination of the \vec{x}_{i}. If \vec{x} = t_{1}\vec{x}_{1} + t_{2}\vec{x}_{2} + \dots + t_{k}\vec{x}_{k} and \vec{y} = s_{1}\vec{x}_{1} + s_{2}\vec{x}_{2} + \dots + s_{k}\vec{x}_{k} are in U, then \vec{x} + \vec{y} and a\vec{x} are in U because

        \begin{eqnarray*} \vec{x} + \vec{y} &=& (t_1 + s_1)\vec{x}_1 + (t_2 + s_2)\vec{x}_2 + \dots + (t_k + s_k)\vec{x}_k, \mbox{ and} \\ a\vec{x} &=& (at_1)\vec{x}_1 + (at_2)\vec{x}_2 + \dots + (at_k)\vec{x}_k \end{eqnarray*}

    Finally each \vec{x}_i is in U (for example, \vec{x}_2 = 0\vec{x}_1 + 1\vec{x}_2 + \dots + 0\vec{x}_k) so S1, S2, and S3 are satisfied for U, proving (1).

  2. Let \vec{x} = t_{1}\vec{x}_{1} + t_{2}\vec{x}_{2} + \dots + t_{k}\vec{x}_{k} where the t_{i} are scalars and each \vec{x}_{i} \in W. Then each t_{i}\vec{x}_{i} \in W because W satisfies S3. But then \vec{x} \in W because W satisfies S2 (verify). This proves (2).

Condition (2) in Theorem 5.1.1 can be expressed by saying that \func{span}\{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} is the smallest subspace of \mathbb{R}^n that contains each \vec{x}_{i}. This is useful for showing that two subspaces U and W are equal, since this amounts to showing that both U \subseteq W and W \subseteq U. Here is an example of how it is used.

 

 

Example 5.1.5

If \vec{x} and \vec{y} are in \mathbb{R}^n, show that \func{span}\{\vec{x}, \vec{y}\} = \func{span}\{\vec{x} + \vec{y}, \vec{x} - \vec{y}\}.

Solution:

Since both \vec{x} + \vec{y} and \vec{x} - \vec{y} are in \func{span}\{\vec{x}, \vec{y}\}, Theorem 5.1.1 gives

    \begin{equation*} \func{span}\{\vec{x} + \vec{y}, \vec{x} - \vec{y} \} \subseteq \func{span}\{\vec{x}, \vec{y}\} \end{equation*}

But \vec{x} = \frac{1}{2}(\vec{x} + \vec{y}) + \frac{1}{2}(\vec{x} - \vec{y}) and \vec{y} = \frac{1}{2}(\vec{x} + \vec{y}) - \frac{1}{2}(\vec{x} - \vec{y}) are both in \func{span}\{\vec{x} + \vec{y}, \vec{x} - \vec{y}\}, so

    \begin{equation*} \func{span}\{\vec{x}, \vec{y}\} \subseteq \func{span}\{\vec{x} + \vec{y}, \vec{x} - \vec{y} \} \end{equation*}

again by Theorem 5.1.1. Thus \func{span}\{\vec{x}, \vec{y}\} = \func{span}\{\vec{x} + \vec{y}, \vec{x} - \vec{y}\}, as desired.

 

 

It turns out that many important subspaces are best described by giving a spanning set. Here are three examples, beginning with an important spanning set for \mathbb{R}^n itself. Column j of the n \times n identity matrix I_{n} is denoted \vec{e}_{j} and called the jth coordinate vector in \mathbb{R}^n, and the set \{\vec{e}_{1}, \vec{e}_{2}, \dots, \vec{e}_{n}\} is called the standard basis of \mathbb{R}^n. If
\vec{x} = \left[ \begin{array}{r} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right] is any vector in \mathbb{R}^n, then \vec{x} = x_{1}\vec{e}_{1} + x_{2}\vec{e}_{2} + \dots + x_{n}\vec{e}_{n}, as the reader can verify. This proves:

Example 5.1.6

\mathbb{R}^n = \func{span}\{\vec{e}_{1}, \vec{e}_{2}, \dots, \vec{e}_{n}\} where \vec{e}_{1}, \vec{e}_{2}, \dots, \vec{e}_{n} are the columns of I_{n}.

 

 

If A is an m \times n matrix A, the next two examples show that it is a routine matter to find spanning sets for \func{null} A and \func{im} A.

Example 5.1.7

Given an m \times n matrix A, let \vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k} denote the basic solutions to the system A\vec{x} = \vec{0} given by the gaussian algorithm. Then

    \begin{equation*} \func{null } A = \func{span}\{\vec{x}_1, \vec{x}_2, \dots, \vec{x}_k\} \end{equation*}

Solution:

If \vec{x} \in \func{null} A, then A\vec{x} = \vec{0} so Theorem 1.3.2 shows that \vec{x} is a linear combination of the basic solutions; that is, \func{null} A \subseteq \func{span}\{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\}. On the other hand, if \vec{x} is in \func{span}\{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\}, then \vec{x} = t_{1}\vec{x}_{1} + t_{2}\vec{x}_{2} + \dots + t_{k}\vec{x}_{k} for scalars t_{i}, so

    \begin{equation*} A\vec{x} = t_1A\vec{x}_1 + t_2A\vec{x}_2 + \dots + t_kA\vec{x}_k = t_1\vec{0} + t_2\vec{0} + \dots + t_k\vec{0} = \vec{0} \end{equation*}

This shows that \vec{x} \in \func{null} A, and hence that \func{span}\{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} \subseteq \func{null} A. Thus we have equality.

Example 5.1.8

Let \vec{c}_{1}, \vec{c}_{2}, \dots, \vec{c}_{n} denote the columns of the m \times n matrix A. Then

    \begin{equation*} \func{im } A = \func{span}\{\vec{c}_1, \vec{c}_2, \dots, \vec{c}_n\} \end{equation*}

Solution:

If \{\vec{e}_{1}, \vec{e}_{2}, \dots, \vec{e}_{n}\} is the standard basis of \mathbb{R}^n, observe that

    \begin{equation*} \left[ \begin{array}{cccc} A\vec{e}_1 & A\vec{e}_2 & \cdots & A\vec{e}_n \end{array}\right] = A \left[ \begin{array}{cccc} \vec{e}_1 & \vec{e}_2 & \cdots & \vec{e}_n \end{array}\right] = AI_n = A = \left[ \begin{array}{cccc} \vec{c}_1 & \vec{c}_2 & \cdots \vec{c}_n \end{array}\right]. \end{equation*}

Hence \vec{c}_{i} = A\vec{e}_{i} is in \func{im} A for each i, so \func{span}\{\vec{c}_{1}, \vec{c}_{2}, \dots, \vec{c}_{n}\} \subseteq \func{ im} A.

Conversely, let \vec{y} be in \func{im} A, say \vec{y} = A\vec{x} for some \vec{x} in \mathbb{R}^n. If
\vec{x} = \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right], then Definition 2.5 gives

    \begin{equation*} \vec{y} = A\vec{x} = x_1\vec{c}_1 + x_2\vec{c}_2 + \dots + x_n\vec{c}_n \mbox{ is in } \func{span} \{\vec{c}_1, \vec{c}_2, \ \dots, \vec{c}_n\} \end{equation*}

This shows that \func{im} A \subseteq \func{span}\{\vec{c}_{1}, \vec{c}_{2}, \dots, \vec{c}_{n}\}, and the result follows.

GeoGebra Exercise: Linear Combination of vectors

https://www.geogebra.org/m/Q4hT3V5N
Please answer these questions after you open the webpage:

1.  Set \vec{OA}=(1,0)and \vec{OB}=(0,1). Set P=(4,5).
2. Click “start animation” to see which linear combination of \vec{OA} and \vec{OB} will produce the vector \vec{OP}.
3. Change \vec{OA}=(-1,2) and \vec{OB}=(2,1). Randomly choose a point P.
4. Click “start animation” to see which linear combination of \vec{OA} and \vec{OB} will produce the vector \vec{OP}. Write it down.
5. What if we set \vec{OA}=(-1,2) and \vec{OB} =(2,-1)? Can you explain what’s happening now? Would you still be able to find a linear combination of \vec{OA} and \vec{OB} to produce a vector \vec{OP}?

 

 

 

5.2 Independence and Dimension

Some spanning sets are better than others. If U = \func{span}\{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} is a subspace of \mathbb{R}^n, then every vector in U can be written as a linear combination of the \vec{x}_{i} in at least one way. Our interest here is in spanning sets where each vector in U has exactly one representation as a linear combination of these vectors.

Linear Independence

Given \vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k} in \mathbb{R}^n, suppose that two linear combinations are equal:

    \begin{equation*} r_1\vec{x}_1 + r_2\vec{x}_2 + \dots + r_k\vec{x}_k = s_1\vec{x}_1 + s_2\vec{x}_2 + \dots + s_k\vec{x}_k \end{equation*}

We are looking for a condition on the set \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} of vectors that guarantees that this representation is unique; that is, r_{i} = s_{i} for each i. Taking all terms to the left side gives

    \begin{equation*} (r_1 - s_1)\vec{x}_1 + (r_2 - s_2)\vec{x}_2 + \dots + (r_k - s_k)\vec{x}_k = \vec{0} \end{equation*}

so the required condition is that this equation forces all the coefficients r_{i} - s_{i} to be zero.

Definition 5.3 Linear Independence in \mathbb{R}^n

We call a set \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} of vectors linearly independent if it satisfies the following condition:

    \begin{equation*} \mbox{If } t_1\vec{x}_1 + t_2\vec{x}_2 + \dots + t_k\vec{x}_k = \vec{0} \mbox{ then } t_1 = t_2 = \dots = t_k = 0 \end{equation*}

Theorem 5.2.1

If \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} is an independent set of vectors in \mathbb{R}^n, then every vector in \func{span}\{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} has a unique representation as a linear combination of the \vec{x}_{i}.

It is useful to state the definition of independence in different language. Let us say that a linear combination vanishes if it equals the zero vector, and
call a linear combination trivial if every coefficient is zero. Then the definition of independence can be compactly stated as follows:

A set of vectors is independent if and only if the only linear combination that vanishes is the trivial one.

Hence we have a procedure for checking that a set of vectors is independent:

Independence Test

To verify that a set \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} of vectors in \mathbb{R}^n is independent, proceed as follows:

  1. Set a linear combination equal to zero: t_{1}\vec{x}_{1} + t_{2}\vec{x}_{2} + \dots + t_{k}\vec{x}_{k} = \vec{0}.
  2. Show that t_i = 0 for each i (that is, the linear combination is trivial).

Of course, if some nontrivial linear combination vanishes, the vectors are not independent.

Example 5.2.1

Determine whether \{(1, 0, -2, 5), (2, 1, 0, -1), (1, 1, 2, 1)\} is independent in \mathbb{R}^4.

Solution:

Suppose a linear combination vanishes:

    \begin{equation*} r(1, 0, -2, 5) + s(2, 1, 0, -1) + t(1, 1, 2, 1) = (0, 0, 0, 0) \end{equation*}

Equating corresponding entries gives a system of four equations:

    \begin{equation*} r + 2s + t = 0, s + t = 0, -2r + 2t = 0, \mbox{ and } 5r -s + t = 0 \end{equation*}

The only solution is the trivial one r = s = t = 0 (please verify), so these vectors are independent by the independence test.

 

Example 5.2.2

Show that the standard basis \{\vec{e}_{1}, \vec{e}_{2}, \dots, \vec{e}_{n}\} of \mathbb{R}^n is independent.

Solution:

The components of t_{1}\vec{e}_{1} + t_{2}\vec{e}_{2} + \dots + t_{n}\vec{e}_{n} are t_{1}, t_{2}, \dots, t_{n}. So the linear combination vanishes if and only if each t_{i} = 0. Hence the independence test applies.

Example 5.2.3

If \{\vec{x}, \vec{y}\} is independent, show that \{2\vec{x} + 3\vec{y}, \vec{x} - 5\vec{y}\} is also independent.

Solution:

If s(2\vec{x} + 3\vec{y}) + t(\vec{x} - 5\vec{y}) = \vec{0}, collect terms to get (2s + t)\vec{x} + (3s - 5t)\vec{y} = \vec{0}. Since \{\vec{x}, \vec{y}\} is independent this combination must be trivial; that is, 2s + t = 0 and 3s - 5t = 0. These equations have only the trivial solution s = t = 0, as required.

 

 

Example 5.2.4

Show that the zero vector in \mathbb{R}^n does not belong to any independent set.

Solution:

No set \{\vec{0}, \vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} of vectors is independent because we have a vanishing, nontrivial linear combination 1 \cdot \vec{0} + 0\vec{x}_{1} + 0\vec{x}_{2} + \dots + 0\vec{x}_{k} = \vec{0}.

Example 5.2.5

Given \vec{x} in \mathbb{R}^n, show that \{\vec{x}\} is independent if and only if \vec{x} \neq \vec{0}.

Solution:

A vanishing linear combination from \{\vec{x}\} takes the form t\vec{x} = \vec{0}, t in \mathbb{R}. This implies that t = 0 because \vec{x} \neq \vec{0}.

Example 5.2.6

Show that the nonzero rows of a row-echelon matrix R are independent.

Solution:

We illustrate the case with 3 leading 1s; the general case is analogous. Suppose R has the form
R = \left[ \begin{array}{rrrrrr} 0 & 1 & * & * & * & * \\ 0 & 0 & 0 & 1 & * & * \\ 0 & 0 & 0 & 0 & 1 & * \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right]
where * indicates a nonspecified number. Let R_{1}, R_{2}, and R_{3} denote the nonzero rows of R. If t_{1}R_{1} + t_{2}R_{2} + t_{3}R_{3} = 0 we show that t_{1} = 0, then t_{2} = 0, and finally t_{3} = 0. The condition t_{1}R_{1} + t_{2}R_{2} + t_{3}R_{3} = 0 becomes

    \begin{equation*} (0, t_1, *, *, *, *) + (0, 0, 0, t_2, *, *) + (0, 0, 0, 0, t_3, *) = (0, 0, 0, 0, 0, 0) \end{equation*}

Equating second entries show that t_{1} = 0, so the condition becomes t_{2}R_{2} + t_{3}R_{3} = 0. Now the same argument shows that t_{2} = 0. Finally, this gives t_{3}R_{3} = 0 and we obtain t_{3} = 0.

 

A set of vectors in \mathbb{R}^n is called linearly dependent (or simply dependent) if it is not linearly independent, equivalently if some nontrivial linear combination vanishes.

Example 5.2.7

If \vec{v} and \vec{w} are nonzero vectors in \mathbb{R}^3, show that \{\vec{v}, \vec{w}\} is dependent if and only if \vec{v} and \vec{w} are parallel.

Solution:

If \vec{v} and \vec{w} are parallel, then one is a scalar multiple of the other, say \vec{v} = a\vec{w} for some scalar a. Then the nontrivial linear combination \vec{v} - a\vec{w} = \vec{0} vanishes, so \{\vec{v}, \vec{w}\} is dependent.

Conversely, if \{\vec{v}, \vec{w}\} is dependent, let s\vec{v} + t\vec{w} = \vec{0} be nontrivial, say s \neq 0. Then \vec{v} = -\frac{t}{s}\vec{w} so \vec{v} and \vec{w} are parallel. A similar argument works if t \neq 0.

With this we can give a geometric description of what it means for a set \{\vec{u}, \vec{v}, \vec{w}\} in \mathbb{R}^3 to be independent. Note that this requirement means that \{\vec{v}, \vec{w}\} is also independent (a\vec{v} + b\vec{w} = \vec{0} means that 0\vec{u} + a\vec{v} + b\vec{w} = \vec{0}), so M = \func{span}\{\vec{v}, \vec{w}\} is the plane containing \vec{v}, \vec{w}, and \vec{0} (see the discussion preceding Example 5.1.4). So we assume that \{\vec{v}, \vec{w}\} is independent in the following example.

Examples 5.2.8

Vectors in R3

Let \vec{u}, \vec{v}, and \vec{w} be nonzero vectors in \mathbb{R}^3 where \{\vec{v}, \vec{w}\} independent. Show that \{\vec{u}, \vec{v}, \vec{w}\} is independent if and only if \vec{u} is not in the plane M = \func{span}\{\vec{v}, \vec{w}\}. This is illustrated in the diagrams.

Solution:

If \{\vec{u}, \vec{v}, \vec{w}\} is independent, suppose \vec{u} is in the plane M = \func{span}\{\vec{v}, \vec{w}\}, say \vec{u} = a\vec{v} + b\vec{w}, where a and b are in \mathbb{R}. Then 1\vec{u} - a\vec{v} - b\vec{w} = \vec{0}, contradicting the independence of \{\vec{u}, \vec{v}, \vec{w}\}.

On the other hand, suppose that \vec{u} is not in M; we must show that \{\vec{u}, \vec{v}, \vec{w}\} is independent. If r\vec{u} + s\vec{v} + t\vec{w} = \vec{0} where r, s, and t are in \mathbb{R}, then r = 0 since otherwise \vec{u} = -\frac{s}{r}\vec{v} + \frac{-t}{r}\vec{w} is in M. But then s\vec{v} + t\vec{w} = \vec{0}, so s = t = 0 by our assumption. This shows that \{\vec{u}, \vec{v}, \vec{w}\} is independent, as required.

By Theorem 2.4.5, the following conditions are equivalent for an n \times n matrix A:

  1. A is invertible.
  2. If A\vec{x} = \vec{0} where \vec{x} is in \mathbb{R}^n, then \vec{x} = \vec{0}.
  3. A\vec{x} = \vec{b} has a solution \vec{x} for every vector \vec{b} in \mathbb{R}^n.

While condition 1 makes no sense if A is not square, conditions 2 and 3 are meaningful for any matrix A and, in fact, are related to independence and spanning. Indeed, if \vec{c}_{1}, \vec{c}_{2}, \dots, \vec{c}_{n} are the columns of A, and if we write
\vec{x} = \left[ \begin{array}{r} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right], then

    \begin{equation*} A\vec{x} = x_1\vec{c}_1 + x_2\vec{c}_2 + \dots + x_n\vec{c}_n \end{equation*}

by Definition 2.5. Hence the definitions of independence and spanning show, respectively, that condition 2 is equivalent to the independence of \{\vec{c}_{1}, \vec{c}_{2}, \dots, \vec{c}_{n}\} and condition 3 is equivalent to the requirement that \func{span}\{\vec{c}_{1}, \vec{c}_{2}, \dots, \vec{c}_{n}\} = \mathbb{R}^m. This discussion is summarized in the following theorem:

Theorem 5.2.2

If A is an m \times n matrix, let \{\vec{c}_{1}, \vec{c}_{2}, \dots, \vec{c}_{n}\} denote the columns of A.

  1. \{\vec{c}_{1}, \vec{c}_{2}, \dots, \vec{c}_{n}\} is independent in \mathbb{R}^m if and only if A\vec{x} = \vec{0}, \vec{x} in \mathbb{R}^n, implies \vec{x} = \vec{0}.
  2. \mathbb{R}^m = \func{span}\{\vec{c}_{1}, \vec{c}_{2}, \dots, \vec{c}_{n}\} if and only if A\vec{x} = \vec{b} has a solution \vec{x} for every vector \vec{b} in \mathbb{R}^m.

For a square matrix A, Theorem 5.2.2 characterizes the invertibility of A in terms of the spanning and independence of its columns (see the discussion preceding Theorem 5.2.2). It is important to be able to discuss these notions for rows. If \vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k} are 1 \times n rows, we define \func{span}\{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} to be the set of all linear combinations of the \vec{x}_{i} (as matrices), and we say that \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} is linearly independent if the only vanishing linear combination is the trivial one (that is, if \{\vec{x}_1^T, \vec{x}_2^T, \dots, \vec{x}_k^T \} is independent in \mathbb{R}^n, as the reader can verify).

Theorem 5.2.3

The following are equivalent for an n \times n matrix A:

  1. A is invertible.
  2. The columns of A are linearly independent.
  3. The columns of A span \mathbb{R}^n.
  4. The rows of A are linearly independent.
  5. The rows of A span the set of all 1 \times n rows.

 

Proof:

Let \vec{c}_{1}, \vec{c}_{2}, \dots, \vec{c}_{n} denote the columns of A.

(1) \Leftrightarrow (2). By Theorem 2.4.5, A is invertible if and only if A\vec{x} = \vec{0} implies \vec{x} = \vec{0}; this holds if and only if \{\vec{c}_{1}, \vec{c}_{2}, \dots, \vec{c}_{n}\} is independent by Theorem 5.2.2.

(1) \Leftrightarrow (3). Again by Theorem 2.4.5,  A is invertible if and only if A\vec{x} = \vec{b} has a solution for every column B in \mathbb{R}^n; this holds if and only if \func{span}\{\vec{c}_{1}, \vec{c}_{2}, \dots, \vec{c}_{n}\} = \mathbb{R}^n by Theorem 5.2.2.

(1) \Leftrightarrow (4). The matrix A is invertible if and only if A^{T} is invertible (by Corollary 2.4.1 to Theorem 2.2.4); this in turn holds if and only if A^{T} has independent columns (by (1) \Leftrightarrow (2)); finally, this last statement holds if and only if A has independent rows (because the rows of A are the transposes of the columns of A^{T}).

(1) \Leftrightarrow (5). The proof is similar to (1) \Leftrightarrow (4).

Example 5.2.9

Show that S = \{(2, -2, 5), (-3, 1, 1), (2, 7, -4)\} is independent in \mathbb{R}^3.

Solution:

Consider the matrix A =  \left[ \begin{array}{rrr} 2 & -2 & 5 \\ -3 & 1 & 1 \\ 2 & 7 & -4 \end{array} \right] with the vectors in S as its rows. A routine computation shows that \func{det} A = -117 \neq 0, so A is invertible. Hence S is independent by Theorem 5.2.3. Note that Theorem 5.2.3 also shows that \mathbb{R}^3 = \func{span} S.

 

 

 

Dimension

It is common geometrical language to say that \mathbb{R}^3 is 3-dimensional, that planes are 2-dimensional and that lines are 1-dimensional. The next theorem is a basic tool for clarifying this idea of “dimension”.

Theorem 5.2.4 Fundamental Theorem

Let U be a subspace of \mathbb{R}^n. If U is spanned by m vectors, and if U contains k linearly independent vectors, then k \leq m

Definition 5.4 Basis of \mathbb{R}^n

If U is a subspace of \mathbb{R}^n, a set \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{m}\} of vectors in U is called a basis of U if it satisfies the following two conditions:

  1. \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{m}\} is linearly independent.
  2. U = \func{span}\{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{m}\}.

Theorem 5.2.5 Invariance Theorem

If \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{m}\} and \{\vec{y}_{1}, \vec{y}_{2}, \dots, \vec{y}_{k}\} are bases of a subspace U of \mathbb{R}^n, then m = k.

Proof:

We have k \leq m by the fundamental theorem because \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{m}\} spans U, and \{\vec{y}_{1}, \vec{y}_{2}, \dots, \vec{y}_{k}\} is independent. Similarly, by interchanging \vec{x}‘s and \vec{y}‘s we get m \leq k. Hence m = k.

 

 

 

Definition 5.5 Dimension of a Subspace of \mathbb{R}^n

If U is a subspace of \mathbb{R}^n and \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{m}\} is any basis
of U, the number, m, of vectors in the basis is called the dimension of U, denoted

    \begin{equation*} \func{dim } U = m \end{equation*}

The importance of the invariance theorem is that the dimension of U can be determined by counting the number of vectors in any basis.

Let \{\vec{e}_{1}, \vec{e}_{2}, \dots, \vec{e}_{n}\} denote the standard basis of \mathbb{R}^n, that is the set of columns of the identity matrix. Then \mathbb{R}^n = \func{span}\{\vec{e}_{1}, \vec{e}_{2}, \dots, \vec{e}_{n}\} by Example 5.1.6, and \{\vec{e}_{1}, \vec{e}_{2}, \dots, \vec{e}_{n}\} is independent by Example 5.2.2. Hence it is indeed a basis of \mathbb{R}^n in the present terminology, and we have

Example 5.2.10

\func{dim}(\mathbb{R}^n) = n and \{\vec{e}_{1}, \vec{e}_{2}, \dots, \vec{e}_{n}\} is a basis.

 

This agrees with our geometric sense that \mathbb{R}^2 is two-dimensional and \mathbb{R}^3 is three-dimensional. It also says that \mathbb{R}^1 = \mathbb{R} is one-dimensional, and \{1\} is a basis. Returning to subspaces of \mathbb{R}^n, we define

    \begin{equation*} \func{dim}\{\vec{0}\} = 0 \end{equation*}

This amounts to saying \{\vec{0}\} has a basis containing no vectors. This makes sense because \vec{0} cannot belong to any independent set.

Example 5.2.11

Let U = \left \{ \left[ \begin{array}{r} r \\ s \\ r \end{array} \right] \mid r, s \mbox{ in } \mathbb{R} \right \}. Show that U is a subspace of \mathbb{R}^3, find a basis, and calculate \func{dim} U.

Solution:

Clearly,
\left[\begin{array}{r} r \\ s \\ r \end{array} \right] = r\vec{u} + s\vec{v} where
\vec{u} =  \left[ \begin{array}{r} 1 \\ 0\\ 1 \end{array} \right] and  \vec{v} = \left[ \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \right]. It follows that U = \func{span}\{\vec{u}, \vec{v}\}, and hence that U is a subspace of \mathbb{R}^3. Moreover, if r\vec{u} + s\vec{v} = \vec{0}, then
\left[ \begin{array}{rrr} r \\ s \\ r \end{array} \right] =  \left[ \begin{array}{rrr} 0 \\ 0 \\ 0 \end{array} \right] so r = s = 0. Hence \{\vec{u}, \vec{v}\} is independent, and so a basis of U. This means \func{dim} U = 2.

 

 

While we have found bases in many subspaces of \mathbb{R}^n, we have not yet shown that every subspace has a basis.

Theorem 5.2.6

Let U \neq \{\normalfont{\vec{0}}\} be a subspace of \mathbb{R}^n. Then:

  1. U has a basis and \func{dim} U \leq n.
  2. Any independent set in U can be enlarged (by adding vectors from the standard basis) to a basis of U.
  3. Any spanning set for U can be cut down (by deleting vectors) to a basis of U.

Example 5.2.3

Find a basis of \mathbb{R}^4 containing S = \{\vec{u}, \vec{v}\} where \vec{u} = (0, 1, 2, 3) and \vec{v} = (2, -1, 0, 1).

Solution:

By Theorem 5.2.6 we can find such a basis by adding vectors from the standard basis of \mathbb{R}^4 to S. If we try \vec{e}_{1} = (1, 0, 0, 0), we find easily that \{\vec{e}_{1}, \vec{u}, \vec{v}\} is independent. Now add another vector from the standard basis, say \vec{e}_{2}.

Again we find that B = \{\vec{e}_{1}, \vec{e}_{2}, \vec{u}, \vec{v}\} is independent. Since B has 4 = \func{dim} \mathbb{R}^4 vectors, then B must span \mathbb{R}^4 by Theorem 5.2.7  below (or simply verify it directly). Hence B is a basis of \mathbb{R}^4.

Theorem 5.2.7

Let U be a subspace of \mathbb{R}^n where \func{dim} U = m and let B = \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{m}\} be a set of m vectors in U. Then B is independent if and only if B spans U.

Proof:

Suppose B is independent. If B does not span U then, by Theorem 5.2.6, B can be enlarged to a basis of U containing more than m vectors. This contradicts the invariance theorem because \func{dim} U = m, so B spans U. Conversely, if B spans U but is not independent, then B can be cut down to a basis of U containing fewer than m vectors, again a contradiction. So B is independent, as required.

Theorem 5.2.8

Let U \subseteq W be subspaces of \mathbb{R}^n. Then:

  1. \func{dim} U \leq \func{dim} W.
  2. If \func{dim} U = \func{dim} W, then U = W.

Proof:

Write \func{dim} W = k, and let B be a basis of U.

  1. If \func{dim} U > k, then B is an independent set in W containing more than k vectors, contradicting the fundamental theorem. So \func{dim} U \leq k = \func{dim} W.
  2. If \func{dim} U = k, then B is an independent set in W containing k = \func{dim} W vectors, so B spans W by Theorem~??. Hence W = \func{span} B = U, proving (2).

It follows from Theorem 5.2.8 that if U is a subspace of \mathbb{R}^n, then \func{dim} U is one of the integers 0, 1, 2, \dots, n, and that:

    \begin{equation*} \begin{array}{l} \func{dim } U = 0 \quad \mbox{ if and only if } \quad U = \{\vec{0}\}, \\ \func{dim } U = n \quad \mbox{ if and only if } \quad U = \mathbb{R}^n \end{array} \end{equation*}

The other subspaces of \mathbb{R}^n are called proper. The following example uses Theorem 5.2.8 to show that the proper subspaces of \mathbb{R}^2 are the lines through the origin, while the proper subspaces of \mathbb{R}^3 are the lines and planes through the origin.

Example 5.2.14

  1. If U is a subspace of \mathbb{R}^2 or \mathbb{R}^3, then \func{dim} U = 1 if and only if U is a line through the origin.
  2. If U is a subspace of \mathbb{R}^3, then \func{dim} U = 2 if and only if U is a plane through the origin.

Solution:

  1. Since \func{dim} U = 1, let \{\vec{u}\} be a basis of U. Then U = \func{span}\{\vec{u}\} = \{t\vec{u} \mid t \mbox{ in }\mathbb{R}\}, so U is the line through the origin with direction vector \vec{u}. Conversely each line L with direction vector \vec{d} \neq \vec{0} has the form L = \{t\vec{d} \mid t \mbox{ in }\mathbb{R}\}. Hence \{\vec{d}\} is a basis of U, so U has dimension 1.
  2. If U \subseteq \mathbb{R}^3 has dimension 2, let \{\vec{v}, \vec{w}\} be a basis of U. Then \vec{v} and \vec{w} are not parallel (by Example 5.2.7) so \vec{n} = \vec{v} \times \vec{w} \neq \vec{0}. Let P = \{\vec{x} in \mathbb{R}^3 \mid \vec{n} \cdot \vec{x} = 0\} denote the plane through the origin with normal \vec{n}. Then P is a subspace of \mathbb{R}^3 (Example 5.1.1) and both \vec{v} and \vec{w} lie in P (they are orthogonal to \vec{n}), so U = \func{span}\{\vec{v}, \vec{w}\} \subseteq P by Theorem 5.1.1. Hence

    \begin{equation*} U \subseteq P \subseteq \mathbb{R}^3 \end{equation*}

Since \func{dim} U = 2 and \func{dim}(\mathbb{R}^3) = 3, it follows from Theorem 5.2.8 that \func{dim} P = 2 or 3, whence P = U or \mathbb{R}^3. But P \neq \mathbb{R}^3 (for example, \vec{n} is not in P) and so U = P is a plane through the origin.

Conversely, if U is a plane through the origin, then \func{dim} U = 0, 1, 2, or 3 by Theorem 5.2.8. But \func{dim} U \neq 0 or 3 because U \neq \{\vec{0}\} and U \neq \mathbb{R}^3, and \func{dim} U \neq 1 by (1). So \func{dim} U = 2.

 

 

 

 

5.3 Orthogonality

Length and orthogonality are basic concepts in geometry and, in \mathbb{R}^2 and \mathbb{R}^3, they both can be defined using the dot product. In this section we extend the dot product to vectors in \mathbb{R}^n, and so endow \mathbb{R}^n with euclidean geometry. We then introduce the idea of an orthogonal basis—one of the most useful concepts in linear algebra, and begin exploring some of its applications.

Dot Product, Length, and Distance

If \vec{x} = (x_{1}, x_{2}, \dots, x_{n}) and \vec{y} =(y_{1}, y_{2}, \dots, y_{n}) are two n-tuples in \mathbb{R}^n, recall that their dot product was defined in Section 2.2 as follows:

    \begin{equation*} \vec{x} \cdot \vec{y} = x_1y_1 + x_2y_2 + \dots + x_ny_n \end{equation*}

  Observe that if \vec{x} and \vec{y} are written as columns then \vec{x} \cdot \vec{y} = \vec{x}^{T}\vec{y} is a matrix product (and \vec{x} \cdotprod \vec{y} = \vecc{x}\vec{y}^{T} if they are written as rows). Here \vec{x} \cdot \vec{y} is a 1 \times 1 matrix, which we take to be a number.

Definition Length in \mathbb{R}^n

As in \mathbb{R}^3, the length of the vector is defined by

    \begin{equation*} || \vec{x} || = \sqrt{\vec{x}\cdot\vec{x}} = \sqrt{x_1^2 + x_2^2 + \dots + x_n^2} \end{equation*}

Where \sqrt{(\quad )} indicates the positive square root.

A vector \vec{x} of length 1 is called a unit vector. If \vec{x} \neq \vec{0}, then || \vec{x} ||\neq 0 and it follows easily that
\frac{1}{|| \vec{x} || }\vec{x} is a unit vector (see Theorem 5.3.6 below), a fact that we shall use later.

Example 5.3.1

If \vec{x} = (1, -1, -3, 1) and \vec{y} = (2, 1, 1, 0) in \mathbb{R}^4, then \vec{x} \cdot \vec{y} = 2 - 1 - 3 + 0 = -2 and || \vec{x} || = \sqrt{1 + 1 + 9 + 1} = \sqrt{12} = 2\sqrt{3}. Hence \frac{1}{2\sqrt{3}}\vec{x} is a unit vector; similarly \frac{1}{\sqrt{6}}\vec{y} is a unit vector.

Theorem 5.3.1

Let \vec{x}, \vec{y}, and \vec{z} denote vectors in \mathbb{R}^n. Then:

  1. \vec{x} \cdot \vec{y} = \vec{y} \cdot \vec{x}.
  2. \vec{x} \cdot (\vec{y} + \vec{z}) = \vec{x} \cdot \vec{y} + \vec{x} \cdot \vec{z}.
  3. (a\vec{x}) \cdot \vec{y} = a(\vec{x} \cdot \vec{y}) = \vec{x} \cdot (a\vec{y}) for all scalars a.
  4. || \vec{x}||^{2} = \vec{x} \cdot \vec{x}.
  5. ||\vec{x}||  \geq 0, and || a \vec{x}|| = 0 if and only if \vec{x} = \vec{0}.
  6. || a \vec{x}||  = | a | ||  \vec{x}|| for all scalars a.

Proof:

(1), (2), and (3) follow from matrix arithmetic because \vec{x} \cdot \vec{y} = \vec{x}^{T}\vec{y}; (4) is clear from the definition; and (6) is a routine verification since |a| = \sqrt{a^2}. If \vec{x} = (x_{1}, x_{2}, \dots, x_{n}), then ||\vec{x}||  = \sqrt{x_1^2 + x_2^2 + \dots + x_n^2} so ||\vec{x}||  = 0 if and only if x_1^2 + x_2^2 + \dots + x_n^2 = 0. Since each x_{i} is a real number this happens if and only if x_{i} = 0 for each i; that is, if and only if \vec{x} = \vec{0}. This proves (5).

Because of Theorem 5.3.1, computations with dot products in \mathbb{R}^n are similar to those in \mathbb{R}^3. In particular, the dot product

    \begin{equation*} (\vec{x}_1 + \vec{x}_2 + \dots + \vec{x}_m) \cdot (\vec{y}_1 + \vec{y}_2 + \dots + \vec{y}_k) \end{equation*}

equals the sum of mk terms, \vec{x}_{i} \cdot \vec{y}_{j}, one for each choice of i and j. For example:

    \begin{align*} (3\vec{x} - 4\vec{y}) \cdot (7\vec{x} + 2\vec{y}) & = 21(\vec{x} \cdot  \vec{x}) + 6(\vec{x} \cdot \vec{y}) - 28(\vec{y} \cdot \vec{x}) - 8(\vec{y} \cdot\vec{y}) \\ &= 21 ||\vec{x}||  ^2 - 22(\vec{x} \cdot \vec{y}) - 8||\vec{y}||^2 \end{align*}

holds for all vectors \vec{x} and \vec{y}.

 

 

 

Example 5.3.2

Show that || \vec{x} + \vec{y}|| ^{2} =||\vec{x}|| ^{2} + 2(\vec{x} \cdot \vec{y}) + ||\vec{y}|| ^{2} for any \vec{x} and \vec{y} in \mathbb{R}^n.

Solution:

Using Theorem 5.3.1 several times:

    \begin{align*} || \vec{x} + \vec{y}||^2 &= (\vec{x} + \vec{y}) \cdot (\vec{x} + \vec{y}) = \vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{y} + \vec{y} \cdot \vec{x} + \vec{y} \cdot \vec{y} \\ &= ||\vec{x}||^2 + 2(\vec{x} \cdot \vec{y}) + ||\vec{y}||^2 \end{align*}

 

Example 5.3.3

Suppose that \mathbb{R}^n = \func{span}\{\vec{f}_{1}, \vec{f}_{2}, \dots, \vec{f}_{k}\} for some vectors \vec{f}_{i}. If \vec{x} \cdot \vec{f}_{i} = 0 for each i where \vec{x} is in \mathbb{R}^n, show that \vec{x} = \vec{0}.

Solution:

We show \vec{x} = \vec{0} by showing that ||\vec{x}|| = 0 and using (5) of Theorem 5.3.1. Since the \vec{f}_{i} span \mathbb{R}^n, write \vec{x} = t_{1}\vec{f}_{1} + t_{2}\vec{f}_{2} + \dots + t_{k}\vec{f}_{k} where the t_{i} are in \mathbb{R}. Then

    \begin{align*} ||\vec{x}||^2 &= \vec{x} \cdot \vec{x} = \vec{x} \cdot (t_1\vec{f}_1 + t_2\vec{f}_2 + \dots + t_k\vec{f}_k) \\ &= t_1(\vec{x} \cdot \vec{f}_1) + t_2(\vec{x} \cdot \vec{f}_2) + \dots + t_k(\vec{x} \cdot \vec{f}_k) \\ &= t_1(0) + t_2(0) + \cdots + t_k(0) \\ &= 0 \end{align*}

We saw in Section 4.2 that if \vec{u} and \vet{v} are nonzero vectors in \mathbb{R}^3, then \frac{\vec{u} \cdot \vec{v}}{ ||\vec{u}|| ||\vec{v}||} = \cos\theta where \theta is the angle between \vec{u} and \vec{v}. Since |\cos \theta| \leq 1 for any angle \theta, this shows that |\vec{u} \cdot \vec{v}| \leq ||\vec{u}|| ||\vec{v}||. In this form the result holds in \mathbb{R}^n.

Theorem 5.3.2 Cauchy Inequality

If \vec{x} and \vec{y} are vectors in \mathbb{R}^n, then

    \begin{equation*} | \vec{x} \cdot \vec{y}| \leq ||\vec{x}|| ||\vec{y}|| \end{equation*}

Moreover |\vec{x} \cdot \vec{y}| = ||\vec{x}|| ||\vec{y}|| if and only if one of \vec{x} and \vec{y} is a multiple of the other.

Proof:

The inequality holds if \vec{x} = \vec{0} or \vec{y} = \vec{0} (in fact it is equality). Otherwise, write ||\vec{x}||  = a > 0 and ||\vec{y}|| = b > 0 for convenience. A computation like that preceding Example 5.3.2 gives

(5.1)   \begin{equation*}  ||  b\vec{x} - a\vec{y} || ^2 = 2ab(ab - \vec{x} \cdot \vec{y}) \mbox{ and } ||  b\vec{x} + a\vec{y} || ^2 = 2ab(ab + \vec{x} \cdot \vec{y}) \end{equation*}

It follows that ab - \vec{x} \cdot \vec{y} \geq 0 and ab + \vec{x} \cdot \vec{y} \geq 0, and hence that -ab \leq \vec{x} \cdot \vec{y} \leq ab. Hence |\vec{x} \cdot \vec{y}| \leq ab = ||\vec{x}|| ||\vec{y}||, proving the Cauchy inequality.

If equality holds, then |\vec{x} \cdot \vec{y}| = ab, so \vec{x} \cdot \vec{y} = ab or \vec{x} \cdot \vec{y} = -ab. Hence Equation (5.1) shows that b\vec{x} - a\vec{y} = 0 or b\vec{x} + a\vec{y} = 0, so one of \vec{x} and \vec{y} is a multiple of the other (even if a = 0 or b = 0).

There is an important consequence of the Cauchy inequality. Given \vec{x} and \vec{y} in \mathbb{R}^n, use Example 5.3.2 and the fact that \vec{x} \cdot \vec{y} \leq ||\vec{x}|| ||\vec{y}|| to compute

    \begin{equation*} || \vec{x} + \vec{y}||^2 = ||\vec{x}|| ^2 + 2(\vec{x} \cdot \vec{y}) + ||\vec{y}||^2 \leq ||\vec{x}|| ^2 + 2 ||\vec{x}|| ||\vec{y}||  + ||\vec{y}||^2 = (|| \vect{x} + \vect{y}|| )^2 \end{equation*}

Taking positive square roots gives:

Corollary 5.3.1 Triangle Inequality

If \vec{x} and \vec{y} are vectors in \mathbb{R}^n, then || \vec{x} + \vec{y}|| \leq || \vect{x} || + || \vec{y}||.

The reason for the name comes from the observation that in \mathbb{R}^3 the inequality asserts that the sum of the lengths of two sides of a triangle is not less than the length of the third side.

 

Definition 5.7 Distance in \mathbb{R}^n

If \vec{x} and \vec{y} are two vectors in \mathbb{R}^n, we define the distance d(\vec{x}, \vec{y}) between \vec{x} and \vec{y} by

    \begin{equation*} d(\vec{x},\vec{y}) = || \vect{x} - \vect{y} || \end{equation*}

Theorem 5.3.3

If \vec{x}, \vec{y}, and \vec{z} are three vectors in \mathbb{R}^n we have:

  1. d(\vec{x}, \vec{y}) \geq 0 for all \vec{x} and \vec{y}.
  2. d(\vec{x}, \vec{y}) = 0 if and only if \vec{x} = \vec{y}.
  3. d(\vec{x}, \vec{y}) = d(\vec{y}, \vec{x}) for all \vec{x} and \vec{y} .
  4. d(\vec{x}, \vec{z}) \leq d(\vec{x}, \vec{y}) + d(\vec{y}, \vec{z})for all \vec{x}, \vec{y}, and \vec{z}. \quad Triangle inequality.

Proof:

(1) and (2) restate part (5) of Theorem 5.3.1 because d(\vec{x}, \vec{y}) = || \vec{x} - \vec{y}||, and (3) follows because ||\vec{u}|| = || -\vec{u}|| for every vector \vec{u} in \mathbb{R}^n. To prove (4) use the Corollary to Theorem 5.3.2:

    \begin{align*} d(\vec{x}, \vec{z}) = || \vec{x} - \vec{z} || &= || (\vec{x} - \vec{y}) + (\vec{y} - \vec{z}) || \\ &\leq || (\vec{x} - \vec{y})|| + || (\vec{y} - \vec{z})|| = d(\vec{x}, \vec{y}) + d(\vec{y}, \vec{z}) \end{align*}

Orthogonal Sets and the Expansion Theorem

Definition 5.8 Orthogonal and Orthonormal Sets

We say that two vectors \vec{x} and \vec{y} in \mathbb{R}^n are orthogonal if \vec{x} \cdot \vec{y} = 0, extending the terminology in \mathbb{R}^3 (See Theorem 4.2.3). More generally, a set \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} of vectors in \mathb{R}^n is called an orthogonal set if

    \begin{equation*} \vec{x}_i \cdot \vec{x}_j = 0 \mbox{ for all } i \neq j \quad \mbox{ and } \quad \vect{x}_i \neq \vect{0} \mbox{ for all } i \end{equation*}

Note that \{\vec{x}\} is an orthogonal set if \vec{x} \neq \vec{0}. A set \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} of vectors in \mathbb{R}^n is called orthonormal if it is orthogonal and, in addition, each \vec{x}_{i} is a unit vector:

    \begin{equation*} || \vec{x}_i ||= 1 \mbox{ for each } i. \end{equation*}

 

Example 5.3.4

The standard basis \{\vec{e}_{1}, \vec{e}_{2}, \dots, \vec{e}_{n}\} is an orthonormal set in \mathbb{R}^n.

Example 5.3.5

If \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} is orthogonal, so also is \{a_{1}\vec{x}_{1}, a_{2}\vec{x}_{2}, \dots, a_{k}\vec{x}_{k}\} for any nonzero scalars a_{i}.

If \vec{x} \neq \vec{0}, it follows from item (6) of Theorem 5.3.1 that \frac{1}{|| \vec{x}||}\vec{x} is a unit vector, that is it has length 1.

Definition 5.9 Normalizing an Orthogonal Set

If \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} is an orthogonal set, then \{  \frac{1}{ || \vec{x_1}||}\vec{x}_1, \frac{1}{|| \vec{x_2}||}\vec{x}_2, \cdots,  \frac{1}{|| \vec{x_k}||}\vec{x}_k \} is an orthonormal set, and we say that it is the result of normalizing the orthogonal set \{\vec{x}_{1}, \vec{x}_{2}, \cdots, \vec{x}_{k}\}.

Example 5.3.6

If \vec{f}_1 =  \left[ \begin{array}{r} 1 \\ 1 \\ 1 \\ -1 \end{array} \right], \vec{f}_2 = \left[ \begin{array}{r} 1 \\ 0 \\ 1 \\ 2 \end{array} \right], \vec{f}_3 = \left[ \begin{array}{r} -1 \\ 0 \\ 1 \\ 0 \end{array} \right], and \vec{f}_4 = \left[ \begin{array}{r} -1 \\ 3 \\ -1 \\ 1 \end{array} \right]
then \{ \vec{f}_1, \vec{f}_2, \vec{f}_3, \vec{f}_4 \} is an orthogonal set in \mathbb{R}^4 as is easily verified. After normalizing, the corresponding orthonormal set is \{ \frac{1}{2}\vec{f}_1, \frac{1}{\sqrt{6}}\vec{f}_2, \frac{1}{\sqrt{2}}\vec{f}_3, \frac{1}{2\sqrt{3}}\vec{f}_4 \}

 

 

The most important result about orthogonality is Pythagoras’ theorem. Given orthogonal vectors \vec{v} and \vec{w} in \mathbb{R}^3, it asserts that

    \begin{equation*} || \vect{v} + \vect{w}|| ^{2} = || \vec{v}||^{2} + || \vec{w}||^{2} \end{equation*}

. In this form the result holds for any orthogonal set in \mathbb{R}^n.

Theorem 5.3.4 Pythagoras’ Theorem

If \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} is an orthogonal set in \mathbb{R}^n, then

    \begin{equation*} || \vec{x}_1 + \vec{x}_2 + \dots + \vec{x}_k || ^2 = \vec{x_1}||^2 + \vec{x_2}||^2 + \dots + \vec{x_k}||^2. \end{equation*}

Proof:

The fact that \vec{x}_{i} \cdot\vec{x}_{j} = 0 whenever i \neq j gives

    \begin{align*} || \vec{x}_1 + \vec{x}_2 + \dots + \vec{x}_k ||^2 & = (\vec{x}_1 + \vec{x}_2 + \dots + \vec{x}_k) \cdot (\vec{x}_1 + \vec{x}_2 + \dots + \vec{x}_k) \\ &= (\vec{x}_1 \cdot \vec{x}_1 + \vec{x}_2 \cdot \vec{x}_2 + \dots + \vec{x}_k \cdot \vec{x}_k) + \sum_{i \neq j}\vec{x}_i \cdot \vec{x}_j \\ &= || \vec{x}_1 || ^2 + || \vec{x}_2||^2 + \dots +|| \vec{x}_k|| ^2 + 0 \end{align*}

This is what we wanted.

If \vec{v} and \vec{w} are orthogonal, nonzero vectors in \mathbb{R}^3, then they are certainly not parallel, and so are linearly independent Example 5.2.7. The next theorem gives a far-reaching extension of this observation.

Theorem 5.3.5

Every orthogonal set in \mathbb{R}^n is linearly independent.

Proof:

Let \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} be an orthogonal set in \mathbb{R}^n and suppose a linear combination vanishes, say: t_{1}\vec{x}_{1} + t_{2}\vec{x}_{2} + \dots + t_{k}\vec{x}_{k} = \vec{0}. Then

    \begin{align*} 0 = \vec{x}_1 \cdot \vec{0} &= \vec{x}_1 \cdot (t_1\vec{x}_1 + t_2\vec{x}_2 + \dots + t_k\vec{x}_k) \\ &= t_1(\vec{x}_1 \cdot \vec{x}_1) + t_2(\vec{x}_1 \cdot \vec{x}_2) + \dots + t_k(\vec{x}_1 \cdot \vec{x}_k) \\ &= t_1 || \vec{x}_1 || ^2 +t_2(0) + \dots + t_k(0) \\ &= t_1|| \vec{x}_1 ||^2 \end{align*}

Since || \vec{x}_{1}|| ^{2} \neq 0, this implies that t_{1} = 0. Similarly t_{i} = 0 for each i.

 

Theorem 5.3.6

Let \{\vec{f}_{1}, \vec{f}_{2}, \dots, \vec{f}_{m}\} be an orthogonal basis of a subspace U of \mathbb{R}^n. If \vec{x} is any vector in U, we have

    \begin{equation*} \vec{x} = \left(\frac{\vec{x} \cdot \vec{f}_1}{|| \vec{f}_1|| ^2} \right) \vec{f}_1 + \left(\frac{\vec{x} \cdot \vec{f}_2}{|| \vec{f}_2 ||^2} \right) \vec{f}_1 + \dots + \left(\frac{\vec{x} \cdot \vec{f}_m}{ || \vec{f}_m ||^2} \right) \vec{f}_m \end{equation*}

Proof:

Since \{\vec{f}_{1}, \vec{f}_{2}, \dots, \vec{f}_{m}\} spans U, we have \vec{x} = t_{1}\vec{f}_{1} + t_{2}\vec{f}_{2} + \dots + t_{m}\vec{f}_{m} where the t_{i} are scalars. To find t_{1} we take the dot product of both sides with \vec{f}_{1}:

    \begin{align*} \vec{x} \cdot \vec{f}_1 &= (t_1\vec{f}_1 + t_2\vec{f}_2 + \dots + t_m\vec{f}_m) \cdot \vec{f}_1 \\ &= t_1(\vec{f}_1 \cdot \vec{f}_1) + t_2(\vec{f}_2 \cdot \vec{f}_1) + \dots + t_m(\vec{f}_m \cdot \vec{f}_1) \\ &= t_1|| \vec{f}_1 || ^2 + t_2(0) + \dots + t_m(0) \\ &= t_1|| \vec{f}_1|| ^2 \end{align*}

Since \vec{f}_{1} \neq \vec{0}, this gives t_1 = \frac{\vec{x} \cdot \vec{f}_1}{|| \vec{f}_1 || ^2}. Similarly, t_i = \frac{\vec{x} \cdot \vec{f}_i}{|| \vec{f}_i || ^2} for each i.

The expansion in Theorem 5.3.6 of \vec{x} as a linear combination of the orthogonal basis \{\vec{f}_{1}, \vec{f}_{2}, \dots, \vec{f}_{m}\} is called the Fourier expansion of \vec{x}, and the coefficients t_1 = \frac{\vec{x} \cdot\vec{f}_i}{|| \vect{f}_i ||^2} are called the Fourier coefficients. Note that if \{\vec{f}_{1}, \vec{f}_{2}, \dots, \vec{f}_{m}\} is actually orthonormal, then t_{i} = \vec{x} \cdot \vec{f}_{i} for each i.

Example 5.3.7

Expand \vec{x} = (a, b, c, d) as a linear combination of the orthogonal basis \{\vec{f}_{1}, \vec{f}_{2}, \vec{f}_{3}, \vec{f}_{4}\} of \mathbb{R}^4 given in Example 5.3.6.

Solution:

We have \vec{f}_{1} = (1, 1, 1, -1), \vec{f}_{2} = (1, 0, 1, 2), \vec{f}_{3} = (-1, 0, 1, 0), and \vec{f}_{4} = (-1, 3, -1, 1) so the Fourier coefficients are

    \begin{equation*} \begin{array}{lcl} t_1 = \frac{\vec{x} \cdot \vec{f}_1}{|| \vec{f}_1|| ^2} = \frac{1}{4}(a + b + c + d) & \hspace{3em} & t_3 = \frac{\vec{x} \cdot \vec{f}_3}{|| \vec{f}_3|| ^2} = \frac{1}{2}(-a + c) \\ & \\ t_2 = \frac{\vec{x} \cdot \vec{f}_2}{|| \vec{f}_2|| ^2} = \frac{1}{6}(a + c + 2d) & & t_4 = \frac{\vec{x} \cdot \vec{f}_4}{|| \vec{f}_4|| ^2} = \frac{1}{12}(-a + 3b -c + d) \end{array} \end{equation*}

The reader can verify that indeed \vec{x} = t_{1}\vec{f}_{1} + t_{2}\vec{f}_{2} + t_{3}\vec{f}_{3} + t_{4}\vec{f}_{4}.

5.4 Rank of a Matrix

In this section we use the concept of dimension to clarify the definition of the rank of a matrix given in Section 1.2, and to study its properties. This requires that we deal with rows and columns in the same way. While it has been the custom to write the n-tuples as columns, in this section we will frequently write them as rows. Subspaces, independence, spanning, and dimension are defined for rows using matrix operations, just as for columns. If A is an m \times n matrix, we define:

Definition 5.10 Column and Row Space of a Matrix

The column space,  \func{col} A, of A is the subspace of \mathbb{R}^m spanned by the columns of A.

The row space, \func{row} A, of A is the subspace of \mathbb{R}^n spanned by the rows of A.

Lemma 5.4.1

Let A and B denote m \times n matrices.

  1. If A \to B by elementary row operations, then \func{row} A = \func{row} B.
  2. If A \to B by elementary column operations, then \func{col} A = \func{col} B.

Proof:

We prove (1); the proof of (2) is analogous. It is enough to do it in the case when A \to B by a single row operation. Let R_{1}, R_{2}, \dots, R_{m} denote the rows of A. The row operation A \to B either interchanges two rows, multiplies a row by a nonzero constant, or adds a multiple of a row to a different row. We leave the first two cases to the reader. In the last case, suppose that a times row p is added to row q where p < q. Then the rows of B are R_{1}, \dots, R_{p}, \dots, R_{q} + aR_p, \dots, R_{m}, and Theorem 5.1.1 shows that

    \begin{equation*} \func{span}\{ R_1, \dots, R_p, \dots, R_q, \dots, R_m \} = \func{span}\{ R_1, \dots, R_p, \dots, R_q + aR_p, \dots, R_m \} \end{equation*}

That is, \func{row} A = \func{row} B.

If A is any matrix, we can carry A \to R by elementary row operations where R is a row-echelon matrix. Hence \func{row} A = \func{row} R by Lemma 5.4.1; so the first part of the following result is of interest.

Lemma 5.4.2

If R is a row-echelon matrix, then

  1. The nonzero rows of R are a basis of \func{row} R.
  2. The columns of R containing leading ones are a basis of \func{col} R.

Proof:

The rows of R are independent, and they span \func{row} R by definition. This proves (1).

Let \vec{c}_{j_{1}}, \vec{c}_{j_{2}}, \dots, \vec{c}_{j_{r}} denote the columns of R containing leading 1s. Then \{\vec{c}_{j_{1}}, \vec{c}_{j_{2}}, \dots, \vec{c}_{j_{r}}\} is independent because the leading 1s are in different rows (and have zeros below and to the left of them). Let U denote the subspace of all columns in \mathbb{R}^m in which the last m - r entries are zero. Then \func{dim}U = r (it is just \mathbb{R}^r with extra zeros). Hence the independent set \{\vec{c}_{j_{1}}, \vec{c}_{j_{2}}, \dots, \vec{c}_{j_{r}}\} is a basis of U by Theorem 5.2.7. Since each \vec{c}_{j_{i}} is in \func{col} R, it follows that \func{col} R = U, proving (2).

Let A be any matrix and suppose A is carried to some row-echelon matrix R by row operations. Note that R is not unique. In Section 1.2 we defined the rank of A, denoted \func{rank} A, to be the number of leading 1s in R, that is the number of nonzero rows of R. The fact that this number does not depend on the choice of R was not proved. However part 1 of Lemma 5.4.2 shows that

    \begin{equation*} \func{rank } A = \func{dim} (\func{row } A) \end{equation*}

and hence that \func{rank} A is independent of R.

Lemma 5.4.2 can be used to find bases of subspaces of \mathbb{R}^n (written as rows). Here is an example.

Example 5.4.1

Find a basis of U = \func{span}\{(1, 1, 2, 3), (2, 4, 1, 0), (1, 5, -4, -9)\}.

Solution:

U is the row space of \left[ \begin{array}{rrrr} 1 & 1 & 2 & 3 \\ 2 & 4 & 1 & 0 \\ 1 & 5 & -4&-9 \end{array} \right]. This matrix has row-echelon form \left[ \begin{array}{rrrr} 1 & 1 & 2 & 3 \\ 0 & 1 & -\frac{3}{2} & -3 \\ 0 & 0 & 0 & 0 \end{array} \right], so \{(1, 1, 2, 3), (0, 1, - \frac{3}{2}, -3)\} is basis of U by Lemma 5.4.1.

Note that \{(1, 1, 2, 3), (0, 2, -3, -6)\} is another basis that avoids fractions.

Theorem 5.4.1 Rank Theorem

Let A denote any m \times n matrix of rank r. Then

    \begin{equation*} \func{dim}(\func{col }A ) = \func{dim} (\func{row } A) = r \end{equation*}

Moreover, if A is carried to a row-echelon matrix R by row operations, then

  1. The r nonzero rows of R are a basis of \func{row} A.
  2. If the leading 1s lie in columns j_{1}, j_{2}, \dots, j_{r} of R, then columns j_{1}, j_{2}, \dots, j_{r} of A are a basis of \func{col} A.

Proof:

We have \func{row} A = \func{row} R by Lemma 5.4.1, so (1) follows from 5.4.2. Moreover, R = UA for some invertible matrix U. Now write A = \left[ \begin{array}{cccc} \vec{c}_1 & \vec{c}_2 & \dots & \vec{c}_n \end{array} \right] where \vec{c}_{1}, \vec{c}_{2}, \dots, \vec{c}_{n} are the columns of A. Then

    \begin{equation*} R = UA = U \left[ \begin{array}{cccc} \vec{c}_1 & \vec{c}_2 & \dots & \vec{c}_n \end{array} \right] =  \left[ \begin{array}{cccc} U\vec{c}_1 & U\vec{c}_2 & \cdots & U\vec{c}_n \end{array} \right] \end{equation*}

Thus, in the notation of (2), the set B = \{U\vec{c}_{j_{1}}, U\vec{c}_{j_{2}}, \dots, U\vec{c}_{j_{r}}\} is a basis of \func{col} R by Lemma 5.4.2. So, to prove (2) and the fact that \func{dim}(\func{col} A) = r, it is enough to show that D = \{\vec{c}_{j_{1}}, \vec{c}_{j_{2}}, \dots, \vec{c}_{j_{r}}\} is a basis of \func{col} A. First, D is linearly independent because U is invertible (verify), so we show that, for each j, column \vec{c}_{j} is a linear combination of the \vec{c}_{j_{i}}. But U\vec{c}_{j} is column j of R, and so is a linear combination of the U\vec{c}_{j_{i}}, say U\vec{c}_{j} = a_{1}U\vec{c}_{j_{1}} + a_{2}U\vec{c}_{j_{2}} + \dots + a_{r}U\vec{c}_{j_{r}} where each a_{i} is a real number.

Since U is invertible, it follows that
\vec{c}_{j} = a_{1}\vec{c}_{j_{1}} +a_{2}\vect{c}_{j_{2}} + \dots + a_{r}\vec{c}_{j_{r}} and the proof is complete.

Example 5.4.2

Compute the rank of  A = \left[ \begin{array}{rrrr} 1 & 2 & 2 & -1 \\ 3 & 6 & 5 & 0 \\ 1 & 2 & 1 & 2 \end{array} \right] and find bases for \func{row} A and \func{col} A.

Solution:

The reduction of A to row-echelon form is as follows:

    \begin{equation*} \left[ \begin{array}{rrrr} 1 & 2 & 2 & -1 \\ 3 & 6 & 5 & 0 \\ 1 & 2 & 1 & 2 \end{array} \right]  \rightarrow \left[\begin{array}{rrrr} 1 & 2 & 2 & -1 \\ 0 & 0 & -1 & 3 \\ 0 & 0 & -1 & 3 \end{array} \right] \rightarrow \left[ \begin{array}{rrrr} 1 & 2 & 2 & -1 \\ 0 & 0 & -1 & 3 \\ 0 & 0 & 0 & 0 \end{array} \right] \end{equation*}

Hence \func{rank} A = 2, and \{ \left[ \begin{array}{cccc} 1 & 2 & 2 & -1 \end{array} \right], \left[ \begin{array}{cccc} 0 & 0 & 1 & -3 \end{array} \right] \}
is a basis of \func{row} A by 5.4.2. Since the leading 1s are in columns 1 and 3 of the row-echelon matrix, Theorem 5.4.1 shows that columns 1 and 3 of A are a basis
\left\{ \left[ \begin{array}{r} 1\\ 3\\ 1 \end{array} \right], \left[ \begin{array}{r} 2\\ 5\\ 1 \end{array} \right] \right\} of \func{col} A.

 

Corollary 5.4.1

If A is any matrix, then \func{rank} A = \func{rank}(A^{T}).

 

 

If A is an m \times n matrix, we have \func{col} A \subseteq \mathbb{R}^m and \func{row} A \subseteq \mathbb{R}^n. Hence Theorem 5.2.8 shows that \func{dim}(\func{col} A) \leq \func{dim}(\mathbb{R}^m) = m and \func{dim}(\func{row} A) \leq \func{dim}(\mathbb{R}^n) = n. Thus Theorem 5.4.1 gives:

Corollary 5.4.2

If A is an m \times n matrix, then \func{rank }A \leq m and \func{rank }A \leq n.

 

Corollary 5.4.3

\func{rank} A = \func{rank}(UA) = \func{rank}(AV) whenever U and V are invertible.

Proof:

Lemma 5.4.1 gives \func{rank} A = \func{rank}(UA). Using this and Corollary 5.4.1 we get

    \begin{equation*} \func{rank} (AV) = \func{rank}(AV)^T = \func{rank} (V^TA^T) = \func{rank}(A^T)= \func{rank } A \end{equation*}

 

 

Lemma 5.4.3

Let A, U, and V be matrices of sizes m \times n, p \times m,
and n \times q respectively.

  1. \func{col}(AV) \subseteq \func{col} A, with equality if VV^{\prime}= I_{n} for some V^{\prime}.
  2. \func{row}(UA) \subseteq \func{row} A, with equality if U^{\prime}U=I_{m} for some U^{\prime}.

Proof:

For (1), write V = \left[ \vec{v}_{1}, \vec{v}_{2}, \dots, \vec{v}_{q}\right] where \vec{v}_{j} is column j of V. Then we have  AV = \left[ A\vec{v}_{1}, A\vec{v}_{2}, \dots, A\vec{v}_{q}\right], and each A\vec{v}_{j} is in \func{col} A by Definition 2.4. It follows that \func{col}(AV) \subseteq \func{col} A. If VV^{\prime}=I_{n}, we obtain \func{col} A = \func{col}\left[ (AV)V^{\prime} \right] \subseteq \func{col}(AV) in the same way. This proves (1).

As to (2), we have \func{col}\left[ (UA)^{T}\right] = \func{col}(A^{T}U^{T}) \subseteq \func{col}(A^{T}) by (1), from which \func{row}(UA) \subseteq \func{row} A. If U^{\prime}U=I_{m}, this is equality as in the proof of (1).

Corollary 5.4.4

If A is m \times n and B is n \times m, then \func{rank} AB \leq \func{rank} A and \func{rank} AB \leq \func{rank} B.

Proof:

By Lemma 5.4.3 \func{col}(AB) \subseteq \func{col} A and \func{row}(BA) \subseteq \func{row} A, so Theorem 5.4.1 applies.

In Section 5.1 we discussed two other subspaces associated with an m \times n matrix A: the null space \func{null}(A) and the image space \func{im}(A)

    \begin{equation*} \func{null} (A) = \{\vec{x} \mbox{ in } \mathbb{R}^n \mid A\vec{x} = \vec{0} \} \mbox{ and } \func{im} (A) = \{A\vec{x} \mid \vec{x} \mbox{ in } \mathbb{R}^n \} \end{equation*}

Using rank, there are simple ways to find bases of these spaces. If A has rank r, we have \func{im}(A) = \func{col}(A) by Example 5.1.8, so \func{dim}[\func{im}(A)] = \func{dim}[\func{col}(A)] = r. Hence Theorem 5.4.1 provides a method of finding a basis of \func{im}(A). This is recorded as part (2) of the following theorem.

Theorem 5.4.2

Let A denote an m \times n matrix of rank r. Then

  1. The n - r basic solutions to the system A\vec{x} = \vec{0} provided by the gaussian algorithm are a basis of \func{null}(A), so \func{dim}[\func{null}(A)] = n - r.
  2. Theorem 5.4.1 provides a basis of \func{im}(A) = \func{col}(A), and \func{dim}[\func{im}(A)] = r.

Proof:

It remains to prove (1). We already know (Theorem 2.2.1) that \func{null}(A) is spanned by the n - r basic solutions of A\vec{x} = \vec{0}. Hence using Theorem 5.2.7, it suffices to show that \func{dim}[\func{null}(A)] = n - r. So let \{\vec{x}_{1}, \dots, \vec{x}_{k}\} be a basis of \func{null}(A), and extend it to a basis \{\vec{x}_{1}, \dots, \vec{x}_{k}, \vec{x}_{k+1}, \dots, \vec{x}_{n}\} of \mathbb{R}^n (by Theorem 5.2.6). It is enough to show that \{A\vec{x}_{k+1}, \dots, A\vec{x}_{n}\} is a basis of \func{im}(A); then n - k = r by the above and so k = n - r as required.

Spanning. Choose A\vec{x} in \func{im}(A), \vec{x} in \mathbb{R}^n, and write \vec{x} = a_{1}\vec{x}_{1} + \dots + a_{k}\vec{x}_{k} + a_{k+1}\vec{x}_{k+1} + \dots + a_{n}\vec{x}_{n} where the a_{i} are in \mathbb{R}.
Then A\vec{x} = a_{k+1}A\vec{x}_{k+1} + \dots + a_{n}A\vec{x}_{n} because \{\vec{x}_{1}, \dots, \vec{x}_{k}\} \subseteq \func{null}(A).

Independence. Let t_{k+1}A\vec{x}_{k+1} + \dots + t_{n}A\vec{x}_{n} = \vec{0}, t_{i} in \mathbb{R}. Then t_{k+1}\vec{x}_{k+1} + \dots + t_{n}\vec{x}_{n} is in \func{null} A, so t_{k+1}\vec{x}_{k+1} + \dots + t_{n}\vec{x}_{n} = t_{1}\vec{x}_{1} + \dots + t_{k}\vec{x}_{k} for some t_{1}, \dots, t_{k} in \mathbb{R}. But then the independence of the \vec{x}_{i} shows that t_{i} = 0 for every i.

 

 

Example 5.4.3

If A =  \left \begin{array}{rrrr} 1 & -2 & 1 & 1 \\ -1 & 2 & 0 & 1 \\ 2 & -4 & 1 & 0 \end{array} \right], find bases of \func{null}(A) and \func{im}(A), and so find their dimensions.

Solution:

If \vec{x} is in \func{null}(A), then A\vec{x} = \vec{0}, so \vec{x} is given by solving the system A\vec{x} = \vec{0}. The reduction of the augmented matrix to reduced form is

    \begin{equation*} \left[ \begin{array}{rrrr|r} 1 & -2 & 1 & 1 & 0 \\ -1 & 2 & 0 & 1 & 0 \\ 2 & -4 & 1 & 0 & 0 \end{array} \right]  \rightarrow \left[ \begin{array}{rrrr|r} 1 & -2 & 0 & -1 & 0 \\ 0 & 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right] \end{equation*}

Hence r = \func{rank}(A) = 2. Here, \func{im}(A) = \func{col}(A) has basis
\left\{ \left[ \begin{array}{r} 1\\ -1\\ 2 \end{array} \right], \left[ \begin{array}{r} 1\\ 0\\ 1 \end{array} \right] \right\} by Theorem 5.4.1 because the leading 1s are in columns 1 and 3. In particular, \func{dim}[\func{im}(A)] = 2 = r as in Theorem 5.4.2.

Turning to \func{null}(A), we use gaussian elimination. The leading variables are x_{1} and x_{3}, so the nonleading variables become parameters: x_{2} = s and x_{4} = t. It follows from the reduced matrix that x_{1} = 2s + t and x_{3} = -2t, so the general solution is

    \begin{equation*} \vec{x} = \left[ \begin{array}{r} x_1\\ x_2\\ x_3\\ x_4 \end{array} \right] = \left[ \begin{array}{c} 2s + t\\ s\\ -2t\\ t \end{array} \right] = s\vec{x}_1 + t\vec{x}_2 \mbox{ where } \vec{x}_1 =  \left[ \begin{array}{r} 2\\ 1\\ 0\\ 0 \end{array} \right], \mbox{ and } \vect{x}_2 = \left[ \begin{array}{r} 1\\ 0\\ -2\\ 1 \end{array} \right]. \end{equation*}

Hence \func{null}(A). But \vec{x}_{1} and \vec{x}_{2} are solutions (basic), so

    \begin{equation*} \func{null}(A) = \func{span}\{\vec{x}_1, \vec{x}_2 \} \end{equation*}

However Theorem 5.4.2 asserts that \{\vec{x}_{1}, \vec{x}_{2}\} is a basis of \func{null}(A). (In fact it is easy to verify directly that \{\vec{x}_{1}, \vec{x}_{2}\} is independent in this case.) In particular, \func{dim}[\func{null}(A)] = 2 = n - r.

Let A be an m \times n matrix. Corollary 5.4.2 asserts that \func{rank} A \leq m and \func{rank} A \leq n, and it is natural to ask when these extreme cases arise. If \vec{c}_{1}, \vec{c}_{2}, \dots, \vec{c}_{n} are the columns of A, Theorem 5.2.2 shows that \{\vec{c}_{1}, \vec{c}_{2}, \dots, \vec{c}_{n}\} spans \mathbb{R}^m if and only if the system A\vec{x} = \vec{b} is consistent for every \vec{b} in \mathbb{R}^m, and that \{\vec{c}_{1}, \vec{c}_{2}, \dots, \vec{c}_{n}\} is independent if and only if A\vec{x} = \vec{0}, \vec{x} in \mathbb{R}^n, implies \vec{x} = \vec{0}. The next two useful theorems improve on both these results, and relate them to when the rank of A is n or m.

 

 

 

 

Theorem 5.4.3

The following are equivalent for an m \times n matrix A:

  1. \func{rank} A = n.
  2. The rows of A span \mathbb{R}^n.
  3. The columns of A are linearly independent in \mathbb{R}^m.
  4. The n \times n matrix A^{T}A is invertible.
  5. CA = I_{n} for some n \times m matrix C.
  6. If A\vec{x} = \vec{0}, \vec{x} in \mathbb{R}^n, then \vec{x} = \vect{0}.

Proof:

(1) \Rightarrow (2). We have \func{row} A \subseteq \mathbb{R}^n, and \func{dim}(\func{row} A) = n by (1), so \func{row} A = \mathbb{R}^n by Theorem 5.2.8. This is (2).

(2) \Rightarrow (3). By (2), \func{row} A = \mathbb{R}^n, so \func{rank} A = n. This means \func{dim}(\func{col} A) = n. Since the n columns of A span \func{col} A, they are independent by Theorem 5.2.7.

(3) \Rightarrow (4). If (A^{T}A)\vec{x} = \vec{0}, \vec{x} in \mathbb{R}^n, we show that \vec{x} = \vec{0} (Theorem 2.4.5). We have

    \begin{equation*} || A\vec{x}|| ^2 = (A\vec{x})^TA\vec{x} = \vec{x}^TA^TA\vec{x} = \vec{x}^T\vec{0} = \vec{0} \end{equation*}

Hence A\vec{x} = \vec{0}, so \vec{x} = \vec{0} by (3) and Theorem 5.2.2.

(4) \Rightarrow (5). Given (4), take C = (A^{T}A)^{-1}A^{T}.

(5) \Rightarrow (6). If A\vec{x} = \vec{0}, then left multiplication by C (from (5)) gives \vec{x} = \vec{0}.

(6) \Rightarrow (1). Given (6), the columns of A are independent by Theorem 5.2.2. Hence \func{dim}(\func{col} A) = n, and (1) follows.

 

 

Theorem 5.4.4

The following are equivalent for an m \times n matrix A:

  1. \func{rank} A = m.
  2. The columns of A span \mathbb{R}^m.
  3. The rows of A are linearly independent in \mathbb{R}^n.
  4. The m \times m matrix AA^T is invertible.
  5. AC = I_m for some n \times m matrix C.
  6. The system A\vec{x} = \vec{b} is consistent for every \vec{b} in \mathbb{R}^m.

Proof:

(1) \Rightarrow (2). By (1), \func{dim}(\func{col} A = m, so \func{col} A = \mathbb{R}^m by Theorem 5.2.8.

(2) \Rightarrow (3). By (2), \func{col} A = \mathbb{R}^m, so \func{rank} A = m. This means \func{dim}(\func{row} A) = m. Since the m rows of A span \func{row} A, they are independent by Theorem 5.2.7.

(3) \Rightarrow (4). We have \func{rank} A = m by (3), so the n \times m matrix A^{T} has rank m. Hence applying Theorem 5.4.3 to A^{T} in place of A shows that (A^{T})^{T}A^{T} is invertible, proving (4).

(4) \Rightarrow (5). Given (4), take C = A^{T}(AA_{T})^{-1} in (5).

(5) \Rightarrow (6). Comparing columns in AC = I_{m} gives A\vec{c}_{j} = \vec{e}_{j} for each j, where \vec{c}_{j} and \vec{e}_{j} denote column j of C and I_{m} respectively. Given \vec{b} in \mathbb{R}^m, write \vec{b} = \sum_{j=1}^{m} r_j\vec{e}_j, r_{j} in \mathbb{R}. Then A\vec{x} = \vec{b} holds with \vec{x}= \sum_{j=1}^{m} r_j\vec{c}_j as the reader can verify.

(6) \Rightarrow (1). Given (6), the columns of A span \mathbb{R}^m by Theorem 5.2.2. Thus \func{col} A = \mathbb{R}^m and (1) follows.

 

 

 

5.5 Similarity and Diagonalization

Similar Matrices

Definition 5.11 Similar Matrices

If A and B are n \times n matrices, we say that A and B are similar, and write A \sim B, if B = P^{-1}AP for some invertible matrix P.

Note that A \sim B if and only if B = QAQ^{-1} where Q is invertible (write P^{-1} = Q). The language of similarity is used throughout linear algebra. For example, a matrix A is diagonalizable if and only if it is similar to a diagonal matrix.

If A \sim B, then necessarily B \sim A. To see why, suppose that B = P^{-1}AP. Then A = PBP^{-1} = Q^{-1}BQ where Q = P^{-1} is invertible. This proves the second of the following properties of similarity:

(5.2)   \begin{align*}  & 1.\ A \sim A \textit{\mbox{ for all square matrices }} A.  \\ & 2.\ \textit{\mbox{If }} A \sim B\textit{\mbox{, then }} B \sim A. \\ & 3.\ \textit{\mbox{If }} A \sim B \textit{\mbox{ and }} B \sim A \textit{\mbox{, then }} A \sim C. \end{align*}

These properties are often expressed by saying that the similarity relation \sim is an equivalence relation on the set of n \times n matrices. Here is an example showing how these properties are used.

Example 5.5.1

If A is similar to B and either A or B is diagonalizable, show that the other is also diagonalizable.

Solution:

We have A \sim B. Suppose that A is diagonalizable, say A \sim D where D is diagonal. Since B \sim A by (2) of (5.2), we have B \sim A and A \sim D. Hence B \sim D by (3) of (5.2), so B is diagonalizable too. An analogous argument works if we assume instead that B is diagonalizable.

Similarity is compatible with inverses, transposes, and powers:

    \begin{equation*} \mbox{If } A \sim B\mbox{ then } \quad A^{-1} \sim B^{-1}, \quad A^T \sim B^T,\quad \mbox{ and } \quad A^k \sim B^k\mbox{ for all integers } k \geq 1. \end{equation*}

 

 

Definition 5.12 Trace of a matrix

The trace \func{tr} A of an n \times n matrix A is defined to be the sum of the main diagonal elements of A.

In other words:

    \begin{equation*} \mbox{If } A = \left[ a_{ij}\right],\mbox{ then } \func{tr } A = a_{11} + a_{22} + \dots + a_{nn}. \end{equation*}

It is evident that \func{tr}(A + B) = \func{tr} A + \func{tr} B and that \func{tr}(cA) = c \func{tr} A holds for all n \times n matrices A and B and all scalars c. The following fact is more surprising.

Lemma 5.5.1

Let A and B be n \times n matrices. Then \func{tr}(AB) = \func{tr}(BA).

Proof:

Write A = \left[ a_{ij} \right] and B = \left b_{ij} \right]. For each i, the (i, i)-entry d_{i} of the matrix AB is given as follows: d_{i} = a_{i1}b_{1i} + a_{i2}b_{2i} + \dots + a_{in}b_{ni} = \sum_{j}a_{ij}b_{ji}. Hence

    \begin{equation*} \func{tr}(AB) = d_1 + d_2 + \dots + d_n = \sum_{i}d_i = \sum_{i}\left(\sum_{j}a_{ij}b_{ji}\right) \end{equation*}

Similarly we have \func{tr}(BA) = \sum_{i}(\sum_{j}b_{ij}a_{ji}). Since these two double sums are the same, Lemma 5.5.1 is proved.

Theorem 5.5.1

If A and B are similar n \times n matrices, then A and B have the same determinant, rank, trace, characteristic polynomial, and eigenvalues.

Proof:

Let B = P^{-1}AP for some invertible matrix P. Then we have

    \begin{equation*} \func{det} B = \func{det}(P^{-1}) \func{det} A \func{det} P = \func{det} A\mbox{ because }\func{det}(P^{-1}) = 1/ \func{det} P \end{equation*}

Similarly, \func{rank} B = \func{rank}(P^{-1}AP) = \func{rank} A by Corollary 5.4.2. Next Lemma 5.5.1 gives

    \begin{equation*} \func{tr} (P^{-1}AP) = \func{tr}\left[ P^{-1}(AP)\right] = \func{tr}\left[ (AP)P^{-1}\right] = \func{tr } A \end{equation*}

As to the characteristic polynomial,

    \begin{align*} c_B(x) = \func{det}(xI - B) &= \func{det} \{x(P^{-1}IP) - P^{-1}AP\} \\ &= \func{det} \{ P^{-1}(xI - A)P\} \\ &= \func{det} (xI - A) \\ &= c_A(x) \end{align*}

Finally, this shows that A and B have the same eigenvalues because the eigenvalues of a matrix are the roots of its characteristic polynomial.

 

 

Example 5.5.2

Sharing the five properties in Theorem 5.5.1 does not guarantee that two matrices are similar. The matrices
A = \left[ \begin{array}{rr} 1 & 1 \\ 0 & 1 \end{array} \right] and I = \left[ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right] have the same determinant, rank, trace, characteristic polynomial, and eigenvalues, but they are not similar because P^{-1}IP = I for any invertible matrix P.

 

 

Diagonalization Revisited

Recall that a square matrix A is diagonalizable if there exists an invertible matrix P such that P^{-1}AP = D is a diagonal matrix, that is if A is similar to a diagonal matrix D\index{diagonal matrices}. Unfortunately, not all matrices are diagonalizable, for example  \left[ \begin{array}{rr} 1 & 1 \\ 0 & 1 \end{array} \right]. Determining whether A is diagonalizable is closely related to the eigenvalues and eigenvectors of A. Recall that a number \lambda is called an eigenvalue of A if A\vec{x} = \lambda\vec{x} for some nonzero column \vec{x} in \mathbb{R}^n, and any such nonzero vector \vec{x} is called an eigenvector of A corresponding to \lambda (or simply a \lambda-eigenvector of A). The eigenvalues and eigenvectors of A are closely related to the characteristic polynomial c_{A}(x) of A, defined by

    \begin{equation*} c_A(x) = \func{det} (xI - A) \end{equation*}

If A is n \times n this is a polynomial of degree n, and its relationship to the eigenvalues is given in the following theorem.

Theorem 5.5.2

Let A be an n \times n matrix.

  1. The eigenvalues \lambda of A are the roots of the characteristic polynomial c_{A}(x) of A.
  2. The \lambda-eigenvectors \vec{x} are the nonzero solutions to the homogeneous system

    \begin{equation*} (\lambda I - A)\vec{x} = \vec{0} \end{equation*}

of linear equations with \lambda I - A as coefficient matrix.

The next theorem will show us the condition when a square matrix is diagonalizable.

Theorem 5.5.3

Let A be an n \times n matrix.

  1. A is diagonalizable if and only if \mathbb{R}^n has a basis \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{n}\} consisting of eigenvectors of A.
  2. When this is the case, the matrix P = \left[ \begin{array}{cccc} \vec{x}_{1} & \vec{x}_{2} & \cdots & \vec{x}_{n} \end{array} \right] is invertible and P^{-1}AP = \func{diag}(\lambda_{1}, \lambda_{2}, \dots, \lambda_{n}) where, for each i, \lambda_{i} is the eigenvalue of A corresponding to \vect{x}_{i}.

The next result is a basic tool for determining when a matrix is diagonalizable. It reveals an important connection between eigenvalues and linear independence: Eigenvectors corresponding to distinct eigenvalues are necessarily linearly independent.

Theorem 5.5.4

Let \vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k} be eigenvectors corresponding to distinct eigenvalues \lambda_{1}, \lambda_{2}, \dots, \lambda_{k} of an n \times n matrix A. Then \{\vec{x}_{1}, \vec{x}_{2}, \dots, \vec{x}_{k}\} is a linearly independent set.

Theorem 5.5.5

If A is an n \times n matrix with n distinct eigenvalues, then A is diagonalizable

Example 5.5.4

Show that A =  \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 2 & 3 \\ -1 & 1 & 0 \\ \end{array} \right] is diagonalizable.

Solution:

A routine computation shows that c_{A}(x) = (x - 1)(x - 3)(x + 1) and so has distinct eigenvalues 1, 3, and -1. Hence Theorem 5.5.5 applies.

Definition 5.1.3 Eigenspace of a Matrix

If \lambda is an eigenvalue of an n \times n matrix A, define the eigenspace of A corresponding to \lambda by

    \begin{equation*} E_{\lambda}(A) = \{\vec{x}\mbox{ in } \mathbb{R}^n \mid A\vec{x} = \lambda\vec{x} \} \end{equation*}

 

 

This is a subspace of \mathbb{R}^n and the eigenvectors corresponding to \lambda are just the nonzero vectors in E_{\lambda}(A). In fact E_{\lambda}(A) is the null space of the matrix (\lambda I - A):

    \begin{equation*} E_{\lambda}(A) = \{\vec{x} \mid (\lambda I - A)\vec{x} = \vec{0} \} = \func{null}(\lambda I - A) \end{equation*}

The basic solutions of the homogeneous system (\lambda I - A)\vec{x} = \vec{0} given by the gaussian algorithm form a basis for E_{\lambda}(A). In particular

(5.5)   \begin{equation*} \func{dim } E_\lambda(A) \mbox{ is the number of basic solutions }\vec{x}\mbox{ of } (\lambda I - A)\vec{x} = \vec{0} \end{equation*}

Now recall that the multiplicity of an eigenvalue \lambda of A is the number of times \lambda occurs as a root of the characteristic polynomial c_{A}(x) of A. In other words, the multiplicity of \lambda is the largest integer m \geq 1 such that

    \begin{equation*} c_A(x) = (x - \lambda)^mg(x) \end{equation*}

for some polynomial g(x). Because of (5.5), a square matrix is diagonalizable if and only if the multiplicity of each eigenvalue \lambda equals \func{dim}\left[ E_{\lambda}(A)\right]. We are going to prove this, and the proof requires the following result which is valid for any square matrix, diagonalizable or not.

 

 

Lemma 5.5.3

Let \lambda be an eigenvalue of multiplicity m of a square matrix A. Then \func{dim}\left[ E_{\lambda}(A)\right] \leq m.

It turns out that this characterizes the diagonalizable n \times n matrices A for which c_{A}(x) factors completely over \mathbb{R}. By this we mean that c_{A}(x) = (x - \lambda_{1})(x - \lambda_{2}) \cdots (x - \lambda_{n}), where the \lambda_{i} are real numbers (not necessarily distinct); in other words, every eigenvalue of A is real. This need not happen (consider A = \left[ \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array} \right]), which leads us to the general conclusion regarding when a square matrix is diagonalizable.

Theorem 5.5.6

The following are equivalent for a square matrix A for which c_{A}(x) factors completely.

  1. A is diagonalizable.
  2. \func{dim}[E_{\lambda}(A)] equals the multiplicity of \lambda for every eigenvalue \lambda of the matrix A

 

 

Example 5.5.5

If A = \left[ \begin{array}{rrr} 5 & 8 & 16 \\ 4 & 1 & 8 \\ -4 & -4 & -11 \end{array} \right] and B =\left[ \begin{array}{rrr} 2 & 1 & 1 \\ 2 & 1 & -2 \\ -1 & 0 & -2 \end{array} \right] show that A is diagonalizable but B is not.

Solution:

We have c_{A}(x) = (x + 3)^{2}(x - 1) so the eigenvalues are \lambda_{1} = -3 and \lambda_{2} = 1. The corresponding eigenspaces are E_{\lambda_1}(A) = \func{span}\{\vec{x}_{1}, \vec{x}_{2}\} and E_{\lambda_2}(A) = \func{span}\{\vec{x}_{3}\} where

    \begin{equation*} \vec{x}_1 = \left[ \begin{array}{r} -1\\ 1\\ 0 \end{array} \right], \vec{x}_2 = \left[  \begin{array}{r} -2\\ 0\\ 1 \end{array} \right] , \vec{x}_3 =  \leftB[ \begin{array}{r} 2\\ 1\\ -1 \end{array} \right] \end{equation*}

as the reader can verify. Since \{\vec{x}_{1}, \vec{x}_{2}\} is independent, we have \func{dim}(E_{\lambda_1}(A)) = 2 which is the multiplicity of \lambda_{1}. Similarly, \func{dim}(E_{\lambda_2}(A)) = 1 equals the multiplicity of \lambda_{2}. Hence A is diagonalizable
by 5.5.6, and a diagonalizing matrix is P =  \left[ \begin{array}{ccc} \vec{x}_{1} & \vec{x}_{2} & \vec{x}_{3} \end{array} \right].

Turning to B, c_{B}(x) = (x + 1)^{2}(x - 3) so the eigenvalues are \lambda_{1} = -1 and \lambda_{2} = 3. The corresponding eigenspaces are E_{\lambda_1}(B) =\func{span}\{\vec{y}_{1}\} and E_{\lambda_2}(B) = \func{span}\{\vec{y}_{2}\} where

    \begin{equation*} \vec{y}_1 = \left[ \begin{array}{r} -1\\ 2\\ 1 \end{array} \right], \vec{y}_2 =\left[ \begin{array}{r} 5\\ 6\\ -1 \end{array} \right] \end{equation*}

Here \func{dim}(E_{\lambda_1}(B)) = 1 is smaller than the multiplicity of \lambda_{1}, so the matrix B is not diagonalizable, again by Theorem 5.5.6. The fact that \func{dim}(E_{\lambda_1}(B)) = 1 means that there is no possibility of finding three linearly independent eigenvectors.

 

 

 

 

 

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